Prove that an isometry f maps the circle c with center P and radius r onto the circle c' with center f(P) and radius r. I know proving an isometry maps circles into circles. However, proving an isometry maps circles onto circles is somewhat similar to the proof of circles into circles. Any hints on how to start this proof? Also, if possible, can I assume the inverse of f, knowing that a bijection is into and onto and then apply what I am trying to claim isometry maps circles onto circles? 
 A: This result follows pretty directly from the fact that, in the metric space $(X,d)$, a circle with center $P$ and radius $r$ is defined as the set $C_P^r = \{x\in X:d(x,P)=r\}$ and that isometries preserve distance. The proof has two parts, the details of which I'll hide below:


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*Show that if $f$ is an isometry of $(X,d)$ then $f^{-1}$ is an isometry of $(X,d)$.



 Let $f$ be an isometry of $(X,d)$. Since $f$ is a bijection by definition, it has an inverse $f^{-1}$. To show that $f^{-1}$ is an isometry, we need to show, given any two points $x,y\in X$, that $d(x,y) = d(f^{-1}(x),f^{-1}(y))$. Since $f$ is an isometry, we have that $d(f^{-1}(x),f^{-1}(y)) = d(f(f^{-1}(x)),f(f^{-1}(y))) = d(x,y)$. Thus $f^{-1}$ is an isometry.



*Prove that $f(C_P^r)\subseteq C_{f(P)}^r$.



 For a point $y\in f(C_P^r)$ there exists some $y'\in C_P^r$ such that $y = f(y')$. Then $r = d(y',P) = d(f(y'),f(P)) = d(y,f(P)) = r$, so $y\in C_{f(P)}^r$, proving the result. 



*Lastly, show that $C_{f(P)}^r\subseteq f(C_P^r)$ by using step $(2)$ with $f^{-1}$.



 (You got this one!)

